Answer:
x=0.01457 m
Explanation:
From parallel axis theorem
I=Icm+mh²
where h=x
The rotational inertia about its center of mass is
Icm=mL²/12
where L=1.0 m
Thus T=4.8s we obtain
[tex]T=2\pi \sqrt{\frac{mL^{2}/12+mx^{2} }{mgx} }\\ T=2\pi \sqrt{\frac{L^{2} }{12gx}+x/g }\\ T^{2}=4\pi^{2}(\frac{L^{2} }{12gx}+x/g )/x\\ T^{2}x=\frac{\pi^{2} L^{2} }{3g}+(\frac{4\pi^{2} }{g})x^{2}\\ 0=(\frac{4\pi^{2} }{g} )x^{2}-(T^{2} )x+(\frac{\pi^{2}L^{2} }{3g} )\\ 4.03x^{2}-23.04x+0.335=0[/tex]
After Solving this quadratic we get
x₁=5.702 m
x₂=0.01457 m
One of the solution is an impossible value for x (x=5.70m is greater than L)
So we choose the other one
x=0.01457 m
The distance x at which the hole is located from the 50 cm mark is; x = 15 cm
We are given;
period; T = 4.8s
The formula for period here is;
T = 2π√(I/mgh)
where;
T is period
I is moment of inertia
m is mass
g is acceleration due to gravity
h is the same as the distance x we are looking for
Now, from parallel axis theorem, moment of inertia is;
I = I_cm + mx²
where;
I_cm = ¹/₁₂mL²
where L is generally equal to 1. Thus;
I_cm = ¹/₁₂m
Thus;
I = ¹/₁₂m + mx²
Put ¹/₁₂m + mx² for I in the period equation to arrive at;
T = 2π√[(1/12gx) + (x/g)]
Simplifying this to a quadratic form gives;
(4π²/g)x² - T²x + (π²/3g)
Putting 4.8 for T and 9.8 for g gives;
4.028x² - 23.04x + 0.3357 = 0
Using online quadratic equation solver , we have;
x = 5.705m or 0.015m
Since the hole is at a distance x from the 50 cm mark, then x cannot be 5.705 but will be 0.015.
Thus;
x = 0.015m or 15 cm
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